Proof. A morphism of varieties $V\to W$ over $k$ is unramified if and only if, for every $w$ in $W$, the morphism $V_w\to \operatorname {Spec} k(w)$ is unramified (i.e. étale); see [Sta20, Tag 00UV]. Now, let $s$ be a point of $S$. To show that the finite surjective morphism $Z_s\to X_s$ is ramified, let $d\in D$ be a point lying over $s$. Then, by the definition of the branch locus, $Z_d\to \operatorname {Spec} k(d)$ is ramified. Note that $Z_d = Z_s\times _{X_s} d$ as schemes over $d=\operatorname {Spec} k(d)$. As the fibre of $Z_s\to X_s$ over $d$ is ramified, it follows that $Z_s\to X_s$ is ramified.

Proof. By [Sta20, Tag 04HJ], there is a surjective étale morphism $f:U\to X$ such that $Y_U :=Y\times _{\pi, X, f} U$ has a finite disjoint union decomposition

\[ Y_U = \bigsqcup_j V_j \]

such that each $V_j\to U$ is a closed immersion. Refining this decomposition if necessary, we may assume that each $V_j$ is connected. As $X$ is a geometrically unibranch integral scheme and $U\to X$ is a surjective étale morphism, it follows that $U$ is a disjoint union of integral schemes. Indeed, each connected component of $U$ is reduced because $U \to X$ is étale and $X$ is integral, hence reduced. Furthermore, each connected component of $U$ is irreducible, for two irreducible components would meet at some point $u$, and the local ring at $u$ would have two minimal primes, contradicting the fact that $X$ is geometrically unibranch [Sta20, Tag 06DM]. Let $U_i$ be the connected component containing the image of $V_j \to U$. Then the restriction of the closed immersion $V_j\to U$ to the integral scheme $U_i$ is a dominant closed immersion, as $V_j \hookrightarrow Y_U$ is an open and closed immersion, $Y_U \to U$ is finite surjective and $Y_U$ is equidimensional (because $Y$ is equidimensional). It follows that each non-trivial $V_j\to U_i$ is an isomorphism. This implies that $Y_U\to U$ is étale, so that $Y\to X$ is étale by étale descent.

Proof. Assume that $\pi$ is unramified. Then, because $X$ is a normal scheme, it is geometrically unibranch [Sta20, Tag 0BQ3]. Therefore, it follows from Lemma 2.2 that $\pi$ is étale, as required.

Proof. There is a canonical embedding $\iota :F \hookrightarrow E' \otimes _E F \hookrightarrow \widehat {F_{E'}}$ that is the identity on $E$. The field $\widehat {F_{E'}}$ is normal over $E'$, and contains $F$ (through the embedding $\iota$). Hence, there exists an embedding $\widehat {F}\hookrightarrow \widehat {F_{E'}}$ which restricts to the identity on $E$. Consider the morphism $\varphi : E' \otimes _E \widehat {F} \rightarrow \widehat {F_{E'}}$ which, by the universal property of the tensor product, is induced by the embedding $E'\hookrightarrow E' \otimes _E F \hookrightarrow \widehat {F_{E'}}$ and the natural embedding $\widehat {F} \hookrightarrow \widehat {F_{E'}}$. The field $\widehat {F}$ is generated over $E$ by the roots $\alpha _1,\ldots,\alpha _n$ of a separable polynomial with coefficients in $E$, and these same roots also generate $\widehat {F_{E'}}$ over $E'$. As both $E'$ and $\alpha _1,\ldots,\alpha _n$ are contained in the image of $\varphi$, this shows that $\varphi$ is surjective as desired.

Proof. By [Sta20, Tag 034F], the schemes $Z'$ and $\widehat {Z} \times _X X'$ are normal. In particular, if $Z'$ is connected, then it is integral (as it is connected and normal). Define

\begin{align*} E := k(X), \quad E' := k(X'),\quad F := k(Z). \end{align*}

Note that

\[ k(\widehat{Z})=\widehat{F}, \quad k(Z')=E' \otimes_E F, \quad k(Z'')=\widehat{F_{E'}}. \]

The result now follows from Proposition 2.9 and the universal property of normalization.

Proof. Let $x\in U(k)$ be a $k$-rational point. Then $Z_x\to \{x\} \times Y$ is a finite étale morphism. Let $Y'\to Z_x\to Y$ be the Galois closure of $Z_x\to Y$, and observe, in particular, that $Y' \to Y$ is finite étale. Let $Z' \subset Z\times _{Y} Y'$ be a connected component of the pull-back of $Z\to Y$ along $Y'\to Y$.

There are natural maps $Z' \to Z \to X \times Y \to X$ and $Z' \to Y'$ which induce a morphism $Z' \to X \times Y'$. Let $Z'\to X'\to X$ be the Stein factorization of the composed morphism $Z'\to X\times Y'\to X$, which (together with the obvious map $Z' \to Y'$) gives a natural morphism $Z'\to X'\times Y'$. We claim that this map is an isomorphism.

To prove this, note that $Z'$ is normal, as $Z'\subset Z \times _Y Y'\to Z$ is finite étale and $Z$ is normal. Moreover, the morphism $Z'\to X'\times Y'$ is finite and surjective. By [Reference GrothendieckGro71, Proposition X.1.2], the Stein factorization of $Z'\to X$ is étale over $U\subset X$. By [Reference GrothendieckGro63, Corollaire 7.8.7], over the étale locus the Stein factorization commutes with taking fibres, so the Stein factorization of $Z'_x\to \operatorname {Spec} k(x)$ is given by $\operatorname {Spec} \Gamma (Z'_x, \mathcal {O}_{Z'_x})$. As $Z'_{x} = Z_{x} \times _Y Y'$ is a disjoint union of copies of $Y'$ (as it follows from Remark 2.8 and a standard normalization argument, noting that $Y'\to Z_{x} \to Y$ is the Galois closure of $Z_x \to Y$), we see that $Z'_x \to X'_x \times Y'$ is an isomorphism (as it is a finite surjective morphism between the same number of copies of $Y'$). It follows that $Z'\to X'\times Y'$ is an isomorphism over a dense open subset, hence it is a birational morphism. As $Z'$ and $X'\times Y'$ are integral normal varieties over $k$, it follows as claimed that $Z'\to X'\times Y'$ is an isomorphism by Zariski's main theorem (see [Reference GrothendieckGro61, Corollaire 4.4.9]).

We have thus constructed the desired diagram and shown parts (1) and (3). Parts (4) and (5) are true by construction. As for part (2), we already know that $X' \to X$ is finite (it arises as the finite part of the Stein factorization of $Z' \to X$) and surjective because $Z' \to X$ is. It remains to show that $X' \to X$ is ramified; if it were not, $Z' \to X \times Y' \to X \times Y$ would be étale, hence also $Z' \to Z \to X \times Y$ would be étale. As we already know that $Z' \to Z$ is surjective and étale, by the cancellation property for étale morphisms we would get that $\pi : Z \to X \times Y$ is also étale, a contradiction.

Proof. As $X$ and $Y$ are integral, it follows that $\phi$ is surjective. Let $( \pi _i : Z_i \to Y)_{i=1,\ldots,n}$ be a finite collection of ramified covers over $k$. By base change, we obtain finite surjective morphisms $\varphi _i : Z_i \times _Y X \to X$.

Let $Z'$ be a connected component of $Z_i \times _Y X$. As $Z'\subset Z_i\times _Y X$ is open and closed, the morphism $Z'\to Z_i$ is finite étale, hence surjective (by connectivity of $Z'$ and $Z_i$). It follows that $\dim Z'=\dim Z_i = \dim Y = \dim X$. In particular, because $\varphi _i|_{Z'}:Z'\to X$ is finite, it is also surjective. Furthermore, the morphism $\varphi _i|_{Z'}:Z'\to X$ is ramified. Indeed, because $X$ is smooth, if $Z' \to X$ is unramified, then it is étale (Lemma 2.3). It follows that, as $Z'\to X$ is étale and $X\to Y$ is étale, the composition $Z' \to X \to Y$ is étale, which by the surjectivity of $Z' \to Z_i$ implies that $Z_i \to Y$ is étale, contradicting our assumption that it is ramified.

Let $Z'_{ij}$ be the connected components of $Z_i \times _Y X$. We may apply the weak-Hilbert property of $X$ to the ramified covers $\varphi _{i}|_{Z'_{ij}} : Z'_{ij} \to X$ to see that $X(k) \setminus \bigcup _{i,j} \varphi _i(Z'_{ij}(k))$ is dense in $X$. Applying the surjective morphism $\phi$ to this dense set of rational points we get a dense set of rational points on $Y$. We claim that this dense set is contained in $Y(k)\setminus \bigcup _i \pi _i(Z_i(k))$. Indeed, suppose by contradiction that there is a point $Q \in X(k)\setminus \bigcup _{i,j} \varphi _i(Z'_{ij}(k))$ such that $\phi (Q)\in \pi _i(Z_i(k))$ for some $i$. Then, by the universal property of the product $Z_i \times _Y X$, the point $Q$ belongs to $\varphi _i((Z_i \times _Y X) (k)) \subset \bigcup _{i,j} \varphi _i(Z'_{ij}(k))$, contradicting our assumption.

Proof. Let $( \pi _i : Z_i \to Y)_{i=1,\ldots,n}$ be a finite collection of ramified covers of $Y$ over $k$. To prove the statement, we have to show that $Y(k) \setminus \bigcup _i \pi _i(Z_i(k))$ is Zariski-dense in $Y$. Let $V \subset X$ be the smooth locus of $\phi$, and let $U$ be the image $\phi (V)$. Note that $V$ is a dense open subscheme of $X$, so that $U \subset Y$ is also an open subscheme. Let $\psi :V\to U$ denote the induced (smooth) morphism. As $\phi$ has no multiple fibres in codimension one, the complement of $U$ in $Y$ is of codimension at least two. As $V\xrightarrow {\phi |_{V}}Y$ is smooth, the morphism $Z_i\times _Y V\rightarrow Z_i$ is smooth. In particular, by [Sta20, Tag 034F] and the normality of $Z_i$, the scheme $W_i:=Z_i\times _Y V$ is normal.

We claim that $W_i$ is integral. Since $k$ is of characteristic zero and $X$ is smooth, the generic fibre $X_{ {k(Y)}}$ of $\phi :X\to Y$ is smooth. As $X_{ {k(Y)}}$ is geometrically connected (by assumption), $X_{ {k(Y)}}$ is geometrically integral. It follows that the same is true for the generic fibre of $V \to Y$. In particular, the generic fibre of $W_i=Z_i\times _Y V\to Z_i$, being the base change of the generic fibre of $V \to Y$, is geometrically integral as well. Applying [Reference LiuLiu06, Proposition 3.8] to $(W_i)_{\overline {k}}\to (Z_i)_{\overline {k}}$, we deduce that $W_i$ is integral.

Define $\pi '_i:W_i \rightarrow V$ to be the natural projection. We claim that $\pi '_i$ is ramified. In fact, assume by contradiction that it is not. We have the following commutative diagram, which is Cartesian in the upper left corner.

(6)

As $\pi '_i$ and $\psi$ are both smooth, so is the morphism $W_i \to U$. Moreover, $W_i \to Z_i|_{U}$ is smooth and surjective (being a base change of the smooth surjective morphism $\psi$). Hence, applying [Sta20, Tag 02K5] to $(f,q,p)=(W_i \to Z_i|_{U}, \pi _i|_{U}, W_i \to U)$, we obtain that $\pi _i|_{U}$ is smooth, hence étale, which by the Zariski–Nagata theorem implies that $\pi _i$ is étale: contradiction.

As $X$ has the weak-Hilbert property, by Remark 3.2, the set $S:=V(k)\setminus \bigcup _{i=1}^n \pi '_i(W_i(k))$ is dense in $X$. By the universal property of the fibred product $W_i=Z_i\times _Y V$, following the same argument used at the end of the proof of Proposition 3.4, we conclude that $\psi (S) \subset Y(k) \setminus \bigcup _i \pi _i(Z_i(k))$ is Zariski-dense in $Y$, as required.

Proof. Let $X\to X'\to Y$ be the Stein factorization of $\phi :X\to Y$. Note that $X\to X'$ is a smooth proper morphism and that $X'\to Y$ is finite étale. In particular, because $X\to X'$ has no multiple fibres in codimension one, it follows that $X'$ has the weak-Hilbert property over $k$ by Proposition 3.6. Then, as $X'$ has the weak-Hilbert property over $k$ and $X'\to Y$ is a finite étale morphism of smooth proper varieties over $k$, we conclude that $Y$ has the weak-Hilbert property over $k$ (Proposition 3.4), as required.

Proof. By standard spreading out arguments, we may choose a regular $\mathbb {Z}$-finitely generated integral domain $A\subset k$, a proper model $\mathcal {X}$ for $X$ over $A$, a proper model $\mathcal {Y}$ for $Y$ over $A$, and a finite étale surjective morphism $F:\mathcal {X}\to \mathcal {Y}$ extending $f$. By properness of $\mathcal {Y}$ over $A$, for every $k$-point $y:\operatorname {Spec} k\to Y$, there exist a dense open subscheme $U_y\subset \operatorname {Spec} A$ whose complement in $\operatorname {Spec} A$ is of codimension at least two and a morphism $U_y\to \mathcal {Y}$ extending the morphism $y:\operatorname {Spec} k\to Y$. Pulling back $U_y\to \mathcal {Y}$ along $F:\mathcal {X}\to \mathcal {Y}$, we obtain a finite étale surjective morphism $V_y:= U_y\times _{\mathcal {Y}} \mathcal {X}\to U_y$ of degree $\deg (f)$. By purity of the branch locus [Reference GrothendieckGro71, Théorème X.3.1], the finite étale morphism $V_y\to U_y$ extends to a finite étale morphism $\overline {V_y}\to \operatorname {Spec} A$. Since the set of isomorphism classes of finite étale covers of $\operatorname {Spec} A$ with bounded degree is finite by the Hermite–Minkowski theorem for arithmetic schemes [Reference Harada and HiranouchiHH09], the set of isomorphism classes of the $\overline {V_y}$ (and, thus, $V_y$) appearing above (with $y\in Y(k)$) is finite. In particular, we may choose a finite field extension $L/k$ such that, for all $k$-points $y:\operatorname {Spec} k\to Y$ and every connected (hence, irreducible) component $V'$ of $V_y$, the function field $K(V')$ of $V'$ is contained in $L$. This readily implies that $Y(k)\subset f(X(L))$, as required.

Proof. Let $Z\to S\to X$ be the Stein factorization of the composed map $Z\to X\times Y\to X$ and note that $S$ is a normal geometrically integral variety over $k$. If $S\to X$ is ramified, there is no need to replace $Z\to X\times Y$ by an arithmetic refinement, and we are done. Thus, we may and do assume that $S\to X$ is unramified, hence étale (Lemma 2.3). Note that the finite surjective morphism $Z\to X\times Y$ factors over a finite surjective morphism $Z\to S\times Y$ which is vertically ramified with respect to $S\times Y\to S$ (note that $S \rightarrow X$ is étale). As the geometric fibres of the composed morphism $Z\to S\times Y\to S$ are connected and $S(k)$ is dense, it follows from Lemma 2.17 that there is the following commutative diagram.

Here $S' \to S$ is a ramified cover, $Y' \to Y$ is finite étale and $\psi :Z'\cong S'\times Y'\to Z$ is finite étale. In particular, because $k$ is a finitely generated field of characteristic zero, it follows from the Chevalley–Weil theorem (Theorem 3.8) that there is a finite field extension $L/k$ such that $Z(k)\subset \psi (Z'(L))$.

Consider the induced morphism of Weil restriction of scalars $R_{L/k}(\psi _L):R_{L/k} Z'_L \to R_{L/k} Z_L$; see [Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 7.6]. Let $\Delta :Z\to R_{L/k} Z_L$ be the diagonal morphism. Let $(W_j)_{j\in J}$ be the connected components of $Z\times _{\Delta, R_{L/k} Z_L} R_{L/k} Z'_L$ such that $W_j(k)\neq \emptyset$ and let $f_j:W_j\to Z$ be the natural morphism. As $W_j$ is normal connected and with a $k$-rational point, it is geometrically integral. Moreover, note that, by the defining properties of the Weil restriction of scalars, the finite set of covers $\{f_j:W_j\to Z\}_{j\in J}$ is an arithmetic refinement of $Z$.

After base change to $\overline {k}$, the fibre product $Z\times _{\Delta, R_{L/k} Z_L} R_{L/k} Z'_L$ is given by $Z'_{\overline {k}}\times _{Z_{\overline {k}}} \ldots \times _{Z_{\overline {k}}} Z'_{\overline {k}}$. Hence, the morphism $(W_j)_{\overline {k}} \rightarrow X_{\overline {k}}$ factors as $(W_j)_{\overline {k}} \rightarrow Z'_{\overline {k}} \rightarrow Z_{\overline {k}} \rightarrow X_{\overline {k}}$, with the morphism $(W_j)_{\overline {k}} \rightarrow Z'_{\overline {k}}$ being finite étale. From this it follows that the Stein factorization of $(W_j)_{\overline {k}}\to X_{\overline {k}}$ factors over the Stein factorization of $Z'_{\overline {k}} = S'_{\overline {k}}\times Y'_{\overline {k}}\to X_{\overline {k}}$ which implies that the Stein factorization of $(W_j)_{\overline {k}}\to X_{\overline {k}}$ is ramified over $X_{\overline {k}}$. As Stein factorization commutes with flat base change (in particular, base change to $\overline {k}$ in our case), we deduce that the Stein factorization of $W_j\to X$ is ramified over $X$, as required.

Proof. As $X$ is geometrically connected and $X\to Y$ is surjective, it follows that $Y$ is geometrically connected over $k$.

We now explain how to reduce to the case that $Y$ (hence, $X$) is projective over $k$. To do so, let $Y'\to Y$ be a proper birational surjective morphism with $Y'$ a smooth projective geometrically connected variety over $k$. Define $X':=X\times _Y Y'$ and note that the natural projection $\pi ':X'\to Y'$ is finite étale (surjective), as it is the base change of $\pi :X\to Y$ along $Y'\to Y$. In particular, because $X'$ is finite étale over the smooth variety $Y'$ over $k$, the scheme $X'$ is a smooth projective geometrically connected variety over $L$. As $X'\to X$ is a proper birational surjective morphism, to show that $X$ has the weak-Hilbert property over $L$, by Proposition 3.1, it suffices to show that $X'$ has the weak-Hilbert property over $L$. Finally, because $Y(k)\subset \pi (X(L))$, it readily follows from the universal property of fibre products that $Y'(k)\subset \pi '(X'(L))$. Thus, replacing $X$ by $X'$ and $Y$ by $Y'$ if necessary, we may and do assume that $X$ and $Y$ are projective over $k$.

Now, let $Z$ be a normal proper variety over $L$ and let $Z\to X$ be a ramified cover. As $X$ and $Y_L$ are projective over $L$, we may consider the Weil restrictions $R_{L/k}(Z)\to R_{L/k}(X)\to R_{L/k}(Y_L)$ and the diagonal map $Y \xrightarrow {\Delta } R_{L/k}(Y_L)$. Let $Z' := R_{L/k}(Z) \times _{R_{L/k}(Y_L)} Y$ and write $Z'=\bigcup _i Z'_i$ for its decomposition in irreducible components. We note that, for each $i$, the natural morphism $Z_i'\to Y$ is finite.

We claim that, if the morphism $Z'_i \rightarrow Y$ is surjective, then the composed morphism $\widetilde {Z_i}'\to Y$ is ramified, where $\widetilde {Z_i}'$ is the normalization of $Z_i'$. To prove this, it suffices to show that $\widetilde {Z_i}'_L \rightarrow Y_L$ is ramified. We have the following commutative diagram:

where the composition of the maps in the lower row is the identity. As the composition $(Z'_i)_L \rightarrow Z \rightarrow Y_L$ is finite surjective and $Z$ is connected, we have that $(Z'_i)_L \rightarrow Z$ is surjective. In particular, because $Z \rightarrow Y_L$ (which is equal to the composition $Z \rightarrow X \rightarrow Y_L$) is ramified, it follows that the composition $\widetilde {Z_i}'_L \rightarrow Z \rightarrow Y_L$ is ramified. (Indeed, if it were unramified, then it would be étale by Lemma 2.3 which would contradict the fact that $Z\to Y_L$ is ramified, as $Z$ is normal.)

Now, by the weak-Hilbert property for $Y$ over $k$, there is a Zariski-dense subset $C\subset Y(k)$ of points not lifting to $k$-points in $Z'$. Let $\Omega \subset X(L)$ be the inverse image of $C$ in $X(L)$ via $\pi : X(L) \to Y(L)$. Note that the assumption $Y(k) \subseteq \pi (X(L))$ ensures that $\pi (\Omega )=C$, hence $\Omega$ is a dense subset of $X$. We claim that, for every $c$ in $\Omega$, the fibre $Z_c$ has no $L$-point. Indeed, such an $L$-point would induce a $k$-point of $R_{L/k}(Z)$ and a $k$-point of $Y$ and, thus, a $k$-point of $Z'$ mapping to $\pi (c) \in Y(k)$: contradiction.

Proof. Define $X:=V_L$ and $Y:=V$. As the morphism $\pi :X=V_L\to V=Y$ is finite étale and $Y(k)=V(k)\subset V(L)= \pi (X(L))$, the proposition follows from Lemma 3.14.

Proof. As $f:V\to U$ is a finite étale morphism and $k$ is a finitely generated field of characteristic zero, by the Chevalley–Weil theorem (Theorem 3.8), there is a finite extension $L/k$ such that $U(k)\subset f(V(L))$. Define $X:=V_L$ and $Y=U$, and let $\pi :X\to Y$ be the composed morphism $V_L\to V\to U$. Note that $Y(k)=U(k)\subset f(V(L))=\pi (X(L))$. Thus, by Lemma 3.14, the weak-Hilbert property holds for $X=V_L$ over $L$.